so i want the area in grey like the image shows, calculating it with a double integral, in polar coordinates.
So for the lower half i put:
r = [-2sin theta , 2cos theta]
theta = [-pi/4 , 0]
for the upper half its almost the same.
r = [2sin theta , 2cos theta]
theta = [0 , pi/4]
Then i set up the integral with those limits and r dr dtheta.
The solution according to the person who made the exercise is 2. But i somehow can't get that. Is there anything wrong here or just some mistake on the calculation of the integrals?
EDIT: my upper half's result is 1, so its the lower one that is screwed, unless this is some kind of coincidence.
EDIT2: Sorry to bother. I found the mistake. It was indeed on the lower half. "r" can't take negative values, so it takes the same values as the upper limit. Only the range of theta is different.
Thanks! Maybe this will help anyone anyway.
Ok, i found the solution. It's actually very simple.
The larger triangle is limited laterally by x = 2, the smaller one by x = 1.
x = rcostheta
so rcostheta = 2
this gives, r = 2 / costheta
(To simplify, 1/ costheta = sectheta)
So the values for each variable are
Theta [0, pi / 4]
R [ sectheta, 2sectheta]